Why am I getting an unresolved external from C++/WinRT if it is a header-only C++ library?

Raymond Chen

A customer was getting unresolved external errors from their C++/WinRT code. How is that possible? C++/WinRT is a header-only library; there is no static library that needs to be linked in. I mean, the way to resolve an unresolved external from a header is to link in the associated static or dynamic library, but for header-only libraries, there is no associated anything to link in. So how can something be unresolved?

C++/WinRT is not just one header, but a series of headers. To reduce compile times, the C++/WinRT headers for a particular component are split into multiple layers of headers. The lowest layer provides forward declarations for types, but almost no definitions. As you go higher up the layers, you get a little more: Maybe a class gets a definition, but the methods don’t get implemented until a still higher layer.

When you include a C++/WinRT header file, it includes declarations, definitions, and implementations for everything for which that header file is responsible, as well as its parent namespaces, but no more.


namespace Contoso.Widgets
    runtimeclass Widget
        Contoso.Gadgets.Gadget GetGadget();
        void Reset();

If you include winrt/Contoso.Widgets.h, you get everything you need to create a Widget and call its methods. However, you don’t get everything you need to operate on Gadget objects. You do have enough to call the GetGadget() method, but you can’t do anything with the gadget until you include winrt/Contoso.Gadgets.h.

Normally, when you make this mistake, you get a compile-time error thanks to a trick: Declare the methods as auto. An auto method must be defined before it can be called, and that’s the language feature that the trick relies on.¹

However, this trick doesn’t work for constructors and destructors, since they don’t have return values that can be auto-ized. If you use a constructor for a type without having included the corresponding namespace header file, the code will compile, and you’ll get a linker error later.²


¹ The methods are, however, defined with a return value in the normal way if __INTELLISENSE__ is defined, so that you get more useful feedback from IntelliSense.

² Well, if you’re lucky and some other translation unit included the namespace header file and used the constructor, then you will get a definition, but that’s only because you were relying on the kindness of strangers.


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